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Mathematics 21 Online
OpenStudy (anonymous):

convergance of sum of Ln(n)/n

OpenStudy (anonymous):

the sum diverges actually

OpenStudy (anonymous):

so we would compare it to 1/n

OpenStudy (anonymous):

yes, ln(n)/n > 1/n, since 1/n diverges, by comparison test, ln(n)/n also diverges

OpenStudy (anonymous):

ok so 1/n diverges by the harmonic series test

OpenStudy (anonymous):

yea

OpenStudy (anonymous):

can u explain tht pls

OpenStudy (anonymous):

The comparison test states that, if an > bn for all natural n and sum (bn) diverges, then sum (an) diverges

OpenStudy (anonymous):

is this the same thing for (ln(n))^2/n

OpenStudy (anonymous):

cause (ln(n))^2 > 1^2

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

kk thank u!

OpenStudy (zarkon):

one minor point: your inequality is not true for all natural numbers...so you should point out that \[\ln(n)>1 \text{ for }n\ge3\] which is all you need. you don't really care what happens at the beginning of the sequence, only what happens after a finite sequence of terms.

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